Lemma 13.4.8. Let $\mathcal{D}$ be a pre-triangulated category. Let

\[ (a, b, c) : (X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h') \]

be a morphism of distinguished triangles. If one of the following conditions holds

$\mathop{\mathrm{Hom}}\nolimits (Y, X') = 0$,

$\mathop{\mathrm{Hom}}\nolimits (Z, Y') = 0$,

$\mathop{\mathrm{Hom}}\nolimits (X, X') = \mathop{\mathrm{Hom}}\nolimits (Z, X') = 0$,

$\mathop{\mathrm{Hom}}\nolimits (Z, X') = \mathop{\mathrm{Hom}}\nolimits (Z, Z') = 0$, or

$\mathop{\mathrm{Hom}}\nolimits (X[1], Z') = \mathop{\mathrm{Hom}}\nolimits (Z, X') = 0$

then $b$ is the unique morphism from $Y \to Y'$ such that $(a, b, c)$ is a morphism of triangles.

**Proof.**
If we have a second morphism of triangles $(a, b', c)$ then $(0, b - b', 0)$ is a morphism of triangles. Hence we have to show: the only morphism $b : Y \to Y'$ such that the $X \to Y \to Y'$ and $Y \to Y' \to Z'$ are zero is $0$. We will use Lemma 13.4.2 without further mention. In particular, condition (3) implies (1). Given condition (1) if the composition $b : Y \to Y' \to Z'$ is zero, then $b$ lifts to a morphism $Y \to X'$ which has to be zero. This proves (1).

The proof of (2) and (4) are dual to this argument.

Assume (5). Consider the diagram

\[ \xymatrix{ X \ar[r]_ f \ar[d]^0 & Y \ar[r]_ g \ar[d]^ b & Z \ar[r]_ h \ar[d]^0 \ar@{..>}[ld]^\epsilon & X[1] \ar[d]^0 \\ X' \ar[r]^{f'} & Y' \ar[r]^{g'} & Z' \ar[r]^{h'} & X'[1] } \]

We may choose $\epsilon $ such that $b = g \circ \epsilon $. Then $g' \circ \epsilon \circ g = 0$ which implies that $g' \circ \epsilon = \delta \circ h$ for some $\delta \in \mathop{\mathrm{Hom}}\nolimits (X[1], Z')$. Since $\mathop{\mathrm{Hom}}\nolimits (X[1], Z') = 0$ we conclude that $g' \circ \epsilon = 0$. Hence $\epsilon = f' \circ \gamma $ for some $\gamma \in \mathop{\mathrm{Hom}}\nolimits (Z, X')$. Since $\mathop{\mathrm{Hom}}\nolimits (Z, X') = 0$ we conclude that $\epsilon = 0$ and hence $b = 0$ as desired.
$\square$

## Comments (1)

Comment #5898 by Taro konno on

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